What are the oxidation states for each element in KClO4—>KCl+2O2?

For the following reaction, KClO4—>KCl+2O2, assign oxidation states to each element on each side of the eq.

Reactants:
K= +1
ClO4= -1 From this you can get…
Cl= +7
O= -2

Products:
K= +1
Cl= -1
O= 0 (because it is by itself)

Chlorine is reduced and Oxygen is oxidized (OILRIG- Oxidation is Losing electrons, Reduction is Gaining electrons)

Source(s): AP chem student

Reactants:
K: +1
Cl: +7
O:-2

Products:
K: 1
Cl: -1
O: 0

Oxidation: Oxygen
Reduced: Chlorine

K will always be a +1, and oxygen will always be a -2. It is the chlorine that can have a number of different states. And you need a neutral molecule. But with diatomic oxygen, it is neutral, so one really doesn’t think in terms of assigning a number.

As far as which is oxidized and which is reduced, it has to do with which way the oxidation number moves, and you need to commit that idea to memory.

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